Since you're asked to use Stoke's theorem, I'm interpreting the question to be asking to compute the line integral
[tex]\displaystyle\int_{\mathcal C}\mathbf f(x,y,z)\cdot\mathrm d\mathbf r[/tex]
which by Stoke's theorem is equivalent to the surface integral
[tex]\displaystyle\iint_{\mathcal S}\nabla\times\mathbf f(x,y,z)\cdot\mathrm d\mathbf S[/tex]
where [tex]\mathcal S[/tex] is the positively-oriented surface with boundary [tex]\mathcal C[/tex].
Given [tex]\mathbf f(x,y,z)=2y\,\mathbf i+3z\,\mathbf j+x\,\mathbf k[/tex], we get curl
[tex]\nabla\times\mathbf f=-3\,\mathbf i-\mathbf j-2\,\mathbf k[/tex]
We parameterize the surface [tex]\mathcal S[/tex] by
[tex]\mathbf s(u,v)=5(1-u)(1-v)\,\mathbf i+5u(1-v)\,\mathbf j+5v\,\mathbf k[/tex]
with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Then we take the partial derivatives of [tex]\mathbf s[/tex] and check their cross product:
[tex]\mathbf s_u\times\mathbf s_v=25(1-v)(\mathbf i+\mathbf j+\mathbf k)[/tex]
Now,
[tex]\mathrm d\mathbf S=(\mathbf s_u\times\mathbf s_v)\,\mathrm du\,\mathrm dv[/tex]
so the surface integral reduces to
[tex]\displaystyle-150\int_{v=0}^{v=1}\int_{u=0}^{u=1}(1-v)\,\mathrm du\,\mathrm dv=-75[/tex]
